Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
const int M = 18;
const int K = 10;
int par[N][M];
vector<int> path[N][M];
int n, m, q;
vector<int> adj[N], ids[N];
int c[N];
int d[N];
vector<int> uni(vector<int> a, vector<int> b) {
  vector<int> res;
  int aa = 0, bb = 0;
  while (a.size() > aa && b.size() > bb && res.size() < K) {
    if (a[aa] < b[bb]) {
      res.push_back(a[aa++]);
    } else {
      res.push_back(b[bb++]);
    }
  }
  while (res.size() < K && aa < a.size()) {
    res.push_back(a[aa++]);
  }
  while (res.size() < K && bb < b.size()) {
    res.push_back(b[bb++]);
  }
  return res;
}
void dfs(int v, int p) {
  if (p != -1) d[v] = d[p] + 1;
  par[v][0] = p;
  for (int i = 1; i < M; i++) {
    if (par[v][i - 1] != -1) {
      par[v][i] = par[par[v][i - 1]][i - 1];
    }
  }
  if (p != -1) path[v][0] = ids[p];
  for (int i = 1; i < M; i++) {
    if (par[v][i] != -1) {
      path[v][i] = uni(path[v][i - 1], path[par[v][i - 1]][i - 1]);
    }
  }
  for (int u : adj[v])
    if (u != p) {
      dfs(u, v);
    }
}
vector<int> lca(int v, int u) {
  if (v == u) return ids[v];
  if (d[v] < d[u]) swap(v, u);
  vector<int> res = ids[v];
  for (int i = M - 1; i >= 0; i--) {
    if (par[v][i] != -1 && d[par[v][i]] >= d[u]) {
      res = uni(res, path[v][i]);
      v = par[v][i];
    }
  }
  if (v == u) return res;
  res = uni(res, ids[u]);
  for (int i = M - 1; i >= 0; i--) {
    if (par[v][i] != par[u][i]) {
      res = uni(res, path[v][i]);
      v = par[v][i];
      res = uni(res, path[u][i]);
      u = par[u][i];
    }
  }
  return uni(res, ids[par[v][0]]);
  ;
}
vector<int> get(int v, int u) {
  if (v == u) return ids[v];
  vector<int> res;
  for (int i = M - 1; i >= 0; i--) {
    if (par[v][i] != -1 && d[par[v][i]] >= d[u]) {
      res = uni(res, path[v][i]);
      v = par[v][i];
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", c + i);
    ids[c[i]].push_back(i);
  }
  memset(par, -1, sizeof par);
  dfs(1, -1);
  while (q--) {
    int v, u, a;
    scanf("%d %d %d", &v, &u, &a);
    vector<int> res = lca(v, u);
    printf("%d", min(a, int(res.size())));
    for (int i = 0; i < min(a, int(res.size())); i++) {
      printf(" %d", res[i]);
    }
    printf("\n");
  }
  return 0;
}